MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $V := \bigoplus_{i \in \mathbb{N}}V_i$ and $W := \bigoplus_{i \in \mathbb{N}}W_i$ are $\mathbb{N}$-graded vector spaces. Then their graded tensor power is defined by $V \bigotimes W := \oplus_{n \in \mathbb{N}} \oplus_{i+j=n} V_i \otimes W_j$ and there is a natural isomorphism $\sigma_{V,W}: V \bigotimes W \to W \bigotimes V$ defined on homogeneous elements by the 'sign convention'

$$\sigma_{V,W}(v \otimes w) = (-1)^{\deg(v)*\deg(w)}w \otimes v$$

What is the origin of this natural isomorphism? Is this the only one or are there others defined by other sign conventions?

share|cite|improve this question
This is known as the Koszul sign convention. My guess (and it is only a guess) is that it has its origins in geometry: for a wedge of differential forms we have $\eta\wedge\omega = (-1)^{\operatorname{deg}(\omega)\operatorname{deg}(\eta)}\omega\wedge\eta$, which makes the exterior derivative well-behaved and keeps track of orientations. – Mark Grant May 10 '12 at 11:15
Even more geometrically, the twist isomorphism $S^n \wedge S^k \cong S^k \wedge S^n$ has degree $(-1)^{n k}$ under the identification of both sides with $S^{n+k}$. – Mike Shulman May 10 '12 at 20:39

Is this the only one or are there others defined by other sign conventions?

As has already been mentioned - there are essentially two but the Koszul one is amongst other things, the only one which makes the tensor product of complexes a complex.

Since I have been interested in this topic of late, I would point you towards the following internal (within MathOverflow) references:

share|cite|improve this answer
1+. This answer basically shows that this question is a duplicate. – Martin Brandenburg May 10 '12 at 14:02

If you want to find another $\sigma'_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'_{U,V\otimes W}=\sigma'_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'_{V,W}\sigma'_{W,V}=Id_{V\otimes W}$, instantly shows that $f(1,1)=\pm1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^*(M)$ (over a field), you then can apply the Künneth formula and say $H^*(M\times N)=H^*(M)\otimes H^*(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^*(M)\otimes H^*(N)$, you need those isomorphisms $\sigma$, so that you can do $H^*(M)\otimes H^*(N)\otimes H^*(M)\otimes H^*(N)\to H^*(M)\otimes H^*(M)\otimes H^*(N)\otimes H^*(N)$ and then compute the product in $H^*(M)$ and in $H^*(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.